Chapter 8:  Fluid Statics

 

A. Density
B. Specific Gravity
C. Pressure
D. Pressure in a Confined Fluid
E. Pascal’s Principle
F. Archimedes’ Principle
Problems
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A simple way to classify matter is as either a solid or a fluid.  Solids are objects with a definite size and shape, and over the first seven chapters we developed laws that describe the equilibrium conditions or motion of solid objects.  Fluids, which include liquids and gases, are distinct from solids because of their abilities to flow and change shape.  In this chapter we will use Newton’s laws to develop techniques for describing fluids in equilibrium.  These techniques will be somewhat similar to methods used for solid bodies, but because the nature of a fluid id so much different that a solid we will need to introduce variables which are appropriate for describing fluids.

A.     Density

A substance can be identified by properties such as color, odor, or crystalline structure.  An important distinguishing attribute of any substance is its density.  Density tells us how much of a substance occupies a given volume, and it can be used to identify solids as well as liquids.  For example, we know that 1 kg of lead will occupy less volume than 1 kg of water, which would occupy less volume than 1 kg of air under standard conditions.  Furthermore, we know that 1 m3 of lead would contain more mass that 1 m3 of water, which would contain more mass than 1 m3 of air under standard conditions.

The density of an object is defined to be mass per unit volume.  Or in symbols

                                                                                                                                                                     (8.1)

The two most common units for mass density are grams per cubic centimeter (g/cm3) and kilograms per cubic meter (kg/m3). 

It is common in the British system to make use of the weight density rather than the mass density.  The weight density is defined to be the weight of a unit volume of material, or, in equation form

                                                                                                                                                     (8.2)

Weight density is commonly expressed in units of pounds per cubic foot.

There are a vast array of units in the field of fluids.  The following relations are very useful for converting units.

Volume:

    1 cm3 = 1 millimeter (ml)

1000 ml = 1000 cm3 = 1 liter

      1 m3 = 1,000,000 cm3

      1 m3 = 35.3 ft3

      1 ft3  = 1728 in3

 1 gallon = 231 in3

Density:                 1 gram/cm3 = 1000 kg/m3 = “62.4 lb/ft3

Note:  The quotation marks surrounding 62.4 lb/ft3 are to emphasize that we have equated a mass density and a weight density.  This simply means that an object at the surface of the earth with a mass density of 1000 kg/m3 would have a weight density of 62.4 lb/ft3.

Examples:

  1. How many liters are there in 1 m3?


  2. Determine the density of the 100 gram aluminum block shown below.



    The volume of a block is given by:
    V = length x width x height
    V = 3.8 cm x 3.8 cm x2.6 cm
    V = 37.5 cm3
    We can now use Eq. 8.1 to find the density.



  3. A baseball has a radius of 3.6 cm and a mass of 145 g.  Determine the density of a baseball/
    The volume of a sphere is given by



    Eq. (8.1) then gives



  4. A golf ball has a radius of 2.1 cm and a mass of 45 g.  Determine its density.





    Question:  The density of water is 1 g/cm3.  Does a golf ball or a baseball float in water?

    Question:  What would you estimate your average density to be?

    The following table gives a representative list of densities.

    Substance Density (gm/cm3)
    Water   1.0
    Air    0.0013
    Styrofoam 0.1
    Bone 1.8
    Blood   1.06
    Iron 7.8
    Mercury 13.6
    Silicon 2.4
    Earth 5.6
    Jupiter  1.3


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B.     Specific Gravity

The specific gravity, or relative density, of a substance is defined to be the ratio of the density of the substance to the density of water.

                                                                                                                   (8.3)

Since it is defined as a ratio of densities, specific gravity is a unitless quantity.  Objects with a specific gravity less than one will float in water.  This simply means that it is less dense than water.

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C.     Pressure

Consider the simple example of pushing your thumb against your nose with a force F.  Such an action may be uncomfortable but it is doubtful that you will seriously injure yourself.  Now try pushing against your nose with the same force F, but this time use a sharp nail instead of you thumb.  In each case described the applied force is the same.  However, much more pain is experienced when you use a nail.  The reason why the nail hurts more than your thumb is because the force is exerted over a large area when your thumb is used.  When the nail is employed this same force is exerted over a more restricted area.  Pressure is a measure of the force exerted per unit area.  It is defined as

                               

Or in symbols                                                                                                                                          (8.4)

The units of pressure are a force divided by an area.  As was the case with volume, there are many units of pressure that are commonly used today.  The following list of conversions illustrates some of the more popular units of pressure.

1 N/m2 = 1 Pascal = 0.021 lb/ft2

1 lb/in2 = 1 psi = 144 lb/ft2

1 Atmosphere = 76 cm of mercury = 100,000 N/m2 = 14.7 lb/in2

               

Example:

Let us examine a student with a mass of 70 kg (weight = 700 N = 154 lb).  Calculate the pressure exerted by the student if he

  1. Stands tiptoe on one foot with a contact area of 10 cm2.
  2. Stands on one cross country ski with a contact area of 1000 cm2.
  3. Stands on one ice skate of length 20 cm and width 0.2 cm


  4. We use Eq. (8.4) to determine the pressure



  5. Again use Eq (8.4)



  6. We must first calculate the area of the skates Area = length x width = 0.2 cm = 4 cm2.  Now the pressure is

                               

Note:  In Chapter 3 we discussed coefficients of kinetic friction.  In that chapter you were told that the coefficient of kinetic friction between ice and steel was 0.9.  However, if there is a layer of water between the ice and steel, then the water acts as a lubricant and reduces the coefficient of kinetic friction to 0.01.  It is clear that this lubricant is necessary for ice skating.  The melting point of ice can be lowered if it is subjected to high pressure.  It is known that every 1200 N/cm2 of pressure will lower the melting point of ice by 1˚C.  In the previous example we calculated the pressure exerted by the ice skate to be 175 N/cm2.  This pressure will lower the melting point of ice only a fraction of a degree.  Thus it seems impossible to ice skate!  Can you think of any mistake or incorrect assumption we have made in this problem?  Think about how the blades on an ice skate are shaped.

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D.     Pressure in a Confined Fluid

Shown below is a block of wood that weighs 100 N (22.5 lb) and an aquarium which holds water weighing 100 N also.  The surface areas on the bottom of the block and the bottom of the aquarium are both 0.1 m2.

Because of its weight the block exerts a pressure on the surface on which it rests.  Using Eq. (8.4) we can readily determine this pressure to be 1,000 N/m2.  Clearly, the weight of the water in the aquarium also exerts a pressure of 1,000 N/m2 on the bottom of the aquarium.

There are two important differences between the pressure exerted by the block and the water in the aquarium.  The first is that the water exerts a pressure on the sides of the aquarium and this pressure increases as the depth below the surface increases.  Because of its nature, however, the block exerts no sideways pressure.

The second difference is a matter of how we measure these pressures.  For a solid it is a simple task to measure its weight and contact area.  This is not the case for a fluid.  Suppose you were 20 meters under water.  How would you determine the weight of water above you?  How would you determine the contact area?  Since both of these measurements would be very difficult we need to find an alternative method for determining pressure in a fluid.

The container below holds a liquid of height h and cross sectional area A.  We will attempt to find an expression for the pressure on the bottom of this container due to weight of the fluid above it.  In addition, we want this expression in terms of variables that are easily measured when dealing with fluids.

The definition of pressure is

                               

Where the force on the bottom of the container is just equal to the weight of the fluid above it, F = mg.  Remember now the definition of density, d = m/V, which allows use to write m = dV.  Substitute these into the pressure equation

                               

Now the volume can be expressed as V = height x length x width = hA.  This allows us to write h = V/A.  Substituting we then find

                Or

                                                                                                                                                                  (8.5)

This tells us the pressure due to a liquid of density d of height h.

Examples:

  1. Find the pressure due to 1 meter of water in units of N/m2, Atm, and psi.  This problem is easily solved using Eq. (8.5) and being careful with units.  First solve the problem in MKS units.  Remember the density of water is 1 gram/cm3 = 1000 kg/m3.



    Using the conversions given earlier in the chapter we can find



    and



  2. Find the pressure due to 10 meters of water in atmospheres. 

    This is very simple using the results of the previous example.  At 1 m the pressure was 0.1 Atm.  Hence the pressure at 10 meters is 10 times larger, or 1 Atm.

Notes:

  1. The result of that last example gives us an easy, and important, value to remember.

    1 Atmosphere = Pressure of 10 meters of water

  2. In the previous examples we computed the pressure due only to the water.  The pressure of the atmosphere contributes to make the total pressure greater.  For example, at a depth of 20 meters the total pressure would be 3 Atm (two Atm of pressure from the water and one from the air).

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E.     Pascal’s Principle

Examine the two blocks in contact shown below.  If a force is exerted on block 1 it will exert a force on block 2.

It is easy to see that the force on block 2 is in the same direction as the applied force on block 1.

Now examine what happens when we exert a force on an incompressible fluid.  Shown below are two such examples.

In both cases the pressure is transmitted to every point within the fluid.  They are examples of Pascal’s principle.

Pressure applied to an incompressible confined fluid

is transmitted to every point in the fluid.

Example:  The best known application of Pascal’s principle is the hydraulic lift, shown schematically below. 

If force F1 is applied to the piston of area A1, the pressure in the fluid is increased throughout by F1/A1.  To maintain the liquid equilibrium, the upward pressure of the liquid on the piston of area A2, must be balanced by an equal downward pressure.  We have by Pascal’s principle that

                               

Solving this for F2 we find

                               

If A2 is much larger that A1 then F2 is much larger than F1.  Thus a relatively small force F1 can support (or lift) a very large weight resting on the piston of area A2.  The walls of the container must, of course, also be strong enough to withstand the large pressure without rupturing.  The same hydraulic principle is used in the brake system of a car.

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F.     Archimedes’ Principle

Let us examine an empty plastic milk container, a baseball, and a rock.  You know that if you try to completely submerge the milk container you must push it under the water.  You also realize that the baseball will float, and that the rock, even though it sinks, appears to weigh less than when it is not submerged.  These observations reflect the fact that an upward force helps to support a submerged object.  We call this a buoyant force

The buoyancy principle , first discovered by Archimedes is as follows:

A body partially or wholly immersed in a fluid is buoyed up

by a force equal to the weight if the fluid when it displaces.

Buoyant Force = Weight if Fluid Displaced

                   BF = mFg                                                                                                                                                (8.6)

where mF is the mass of the displaced fluid.

Note:  When solving buoyancy problems it is helpful to remember the definition of density

D = m/V, m = dV, or V = m/d

Examples:  We will consider three different examples where buoyant forces play an important role.

  1. A nearly weightless but rigid 1-liter milk carton is held submerged under water.
    1. What is the volume of the carton in m3?  We know that 1,000 liter  = 1m3.  So

              

    2. What is the volume of water displaced?

      Since the milk carton is completely submerged, the volume of water displaces is equal to the volume of the carton.

      Volume of water displaced = 1 liter = 0.001 m3

    3. What mass of water is displaced?

      Here we use the definition of density and the known density of water (1000 kg/m3).



    4. What weight of water is displaced?

      Now use the definition of weight.

      W = mg = (1kg) (10m/sec2) = 10 N

    5. What force is required to hold the milk carton underwater?

      In this example the force that must be applied is equal and opposite to the upward buoyant force.  Since the buoyant force is simply the weight of the displaced fluid we have

      F = BF = Weight of displaced water = 10 N

  2. A student floats with 95% of his body submerged.  What is her density?

    Since she is floating, this is an equilibrium situation.  Here the weight of the student (Ws) down is balanced by the buoyant force up.

    Ws = BF = mFg

    Once again the problem is to find the mass of the displaced fluid.  Again we can make use of the density equation, mF = dFVF.  This gives

    BF = mFg = dFVFg

    Now let us write the weight of the student in terms of her density and volume.

    Ws = msg = dsVsg

    Equating the weight of the student and buoyant force we have

                   

    Since she is not completely submerged the volume of fluid displaced is less than her volume.  We have



    Hence



  3. A metal object has a mass of 10 g in air and an apparent mass of 6 g in water.

    1. What mass of water does it displace when it is submerged?

      The buoyant force seems to have reduced the mass from 10 g to an apparent mass of 6 g.  As a result, 4 g of water have been displaced.

    2. What volume of water does it displace?

      4 g of water with density 1 g/cm3 have been displaced.  From the definition of density we have





    3. What is the volume of the metal?

      Because it is completely submerged we must have

      Volume of metal = Volume of displaced water = 4 cm3

    4. What is the density of the metal?

      From the density definition



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Summary

Mass Density    Pressure in a Confined Fluid
Weight Density   Pascal’s Principle
Specific Gravity Achimedes’ Principle
Pressure  

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Problems

  1. A gallon is equivalent to 0.134 ft3 or 3800 cm3.  Find the mass of one gallon of water in grams and kilograms, and the weight in pounds.

  2. A steel sphere (V = 4/3 πr3) has a radius of 0.5 cm and a mass of 4.15 grams.  What is its density?

  3. The human body has a density close to that of water (1 gram/cm3 = 1000 kg/m3).  Use this to estimate your volume.

  4. How many people with a volume equal to yours could you stuff into a 1 m3 box?

  5. A baseball player wears a helmet to protect his head when he bats.  The helmet reduced the pressure mainly by spreading the force over a large area.  Assume that the force exerted on the head is spread over an area of 4 cm2 if it is unprotected, but if a helmet is worn this force is over and area of 200 cm2.  Compare the pressure of being hit with a pitched baseball in the head without a helmet.

    Besides spreading the force over a larger area, can you think of any other way in which the helmet reduces the impact of a baseball?

  6. Calculate the pressure on the fluid on a syringe when a nurse applies a force of 40 N to the syringe’s piston of radius 1 cm (A = πr2).

  7. A pickle jar has a lid of area 12 in2.  The pickles are packed under lowered pressure and it is found that a force of 108 lb is required to pull the lid off the jar.  If the outside pressure is one atmosphere (14.7 lb/in2) what is the pressure in the jar?

  8. What is the total pressure 1 mile (1609 meters) below the surface of water?

  9. As bubbles of gas rise in a glass of a carbonated beverage do they become larger, smaller, or remain the same size as they approach the surface?  Why?

  10. The human lungs can operate against a pressure differential of less that one-twentieth of a standard atmosphere (1.05 Atm).  If a diver uses a snorkel, how far below the water level can he breathe?

  11. Find the difference in blood pressure between the feet and the head in a person 1.8 m (6 ft) tall.  The density of blood is 1060 kg/m3.

  12. Why is your blood pressure measure on your upper arm?

  13. Atmospheric pressure is measured by using a barometer.  The barometer is a fairly simple device.  A beaker is filled with mercury and an open glass tube is immersed in it.  A vacuum pump is connected to the end of the tube, all the air is removed, and the end is sealed.  There will then be an unbalanced force in the mercury that tends to push it up the tube.  The mercury will rise to such a height that the pressure due to mercury is equal to the pressure of the atmosphere.  To what height will the mercury rise at atmospheric pressure (1 atmosphere = 100,000 N/m2)?  The density of mercury is 13,600 kg/m3 (13.6 g/cm3).

  14. If a barometer is made with water (density = 1000 kg/m3) instead of mercury, to what height would the water rise at standard atmospheric pressure?

  15. A 2000 lb car is lifted by a manually operated hydraulic lift.  The area of the shaft of the lift is 80 cm2; the area of the piston that forces liquid into the system is 2 cm2.  What force must be exerted on this piston to lift the car?

  16. Find the force required to push a nearly weightless, but rigid, 2 liter milk carton under water.

  17. A student has a density of 1050 kg/m3 (he would sink in ordinary water).  His mass is 70 kg.
    1. What is his weight?
    2. What is his volume?
    3. If he is submerged in water what volume of water would be displace?
    4. Find the mass of the displaced water.
    5. What is the buoyant force on him?
    6. What is his apparent weight in water?

  18. We are all familiar with the expression “only the tip of the iceberg”.  Ice has a density of 0.92 g/cm3.  If an iceberg is floating in water (density = 1.00 g/cm3) what fraction of an iceberg is submerged?  The density of salt water is slightly larger than pure water.  Will more or less of the iceberg be submerged in seawater?

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